Spacecraft and Aircraft Dynamics - Lecture 6: The Orbital Plane Lecture 6: The Orbital Plane
Introduction
The Orbital Plane
• The line of nodes
• How to construct all orbital elements from ~r and ~v
• A Numerical Illustration
The Orbital Elements 2D orbits
So far, all orbits are parameterized by 3 parameters
• semimajor axis, a
• a and e define the geometry of the orbit.
• f describes the position within the orbit (a proxy for time).
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The Orbital Elements
Note: We have shown how to use a, e and f to find the scalars r and v.
Question: How do we find the vectors ~r and ~v?
Answer: We have to determine how the orbit is oriented in space.
• Orientation is determined by vectors ~e and ~h.
• We need 3 new orbital elements I Orientation can be determined by 3 rotations.
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The Coordinate System Earth-Centered Inertial (ECI)
Question: How do we find the vectors ~r and ~v? Response: In which coordinate system??
• The origin is the center of the earth
• We need to define the x, y, and z vectors.
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ECI: The Equatorial Plane Defining the z vector
• The z vector is defined to be the vector parallel to the axis of rotation of the earth.
• Can apply to other planets
• Does not apply to Heliocentric Coordinates
Definition 1.
The Equatorial Plane is the set of vectors normal to the axis of rotation.
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The Ecliptic Plane Heliocentric Coordinates
The rotation vector of the sun is unreliable.
• In heliocentric coordinates, the z vector is normal to the ecliptic plane.
Definition 2.
The Ecliptic Plane is the orbital plane of the earth in motion about the sun.
• From the earth, the ecliptic plane is defined by the apparent motion of the sun about the earth.
I Determined by the location of eclipses (hence the name).
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The Ecliptic Plane
Definition 3.
The Inclination to the Ecliptic is the angle between the equatorial and ecliptic planes.
Currently, the inclination to the ecliptic is 23.5 deg. M. Peet Lecture 6: Spacecraft Dynamics 8 / 29
ECI: The First Point of Aries
• To define the ECI coordinate system, we define x axis in the equatorial plane.
• The y axis is then given by the right-hand rule.
A convenient choice for x is the intersection of the equatorial and ecliptic planes. But there are two such points, at the two equinoxes (vernal and autumnal)
The First Point of Aries is the earth-sun vector at the vernal equinox.
Question Does the FPOA lie at the ascending or descending node? M. Peet Lecture 6: Spacecraft Dynamics 9 / 29
ECI: The First Point in Aries
• The First Point of Aries is so named because this direction used to point towards the Constellation Aries.
• Precession of the earth’s rotation vector means the FPOA now actually points toward Pisces.
• Since Motion of the FPOA is caused be precession, its motion is Periodic, not Secular.
I The Period is about 26,000 years.
• The Coordinate System is not truly inertial.
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Summary: The ECI frame
• y - Right Hand Rule
Because the FPOA migrates with time, positions given in ECI must be referenced to a year
• J2000 - frame as defined at 12:00 TT on Jan 1, 2000.
• TOD - True of Data: date is listed explicitly.
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Other Reference Frames
• Earth Centered Earth Fixed
• Equinoctal
We will return to some of these frames when necessary. M. Peet Lecture 6: Spacecraft Dynamics 12 / 29
Orbital Elements
Now that we have our coordinate system,
Question:, Suppose we are given ~r and ~v in the ECI frame. How to describe the orientation of the orbit?
Answer: 3 new orbital elements.
• Inclination • Right Ascension • Argument of Periapse
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The Orbital Plane Inclination, i
Angle the orbital plane makes with the reference plane. The orbit is
• Prograde if 0 < i < 90.
• Retrograde if 90 < i < 180.
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The Orbital Plane Inclination, i
Inclination can be found from ~h as
~h · z = h cos i.
• If ~h is defined in ECI, then i = cos−1 h3
h .
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The Line Of Nodes
An important vector in defining the orbit is the line of nodes.
Definition 4.
The Line of Nodes is the vector pointing to where the satellite crosses the equatorial plane from the southern to northern hemisphere.
~n = z × ~h
The Line Of Nodes
• Lies at the intersection of the equatorial and orbital planes.
• Points toward the Ascending Node.
• Zero for equatorial orbits (i = 0).
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The Orbital Plane Right Ascension of Ascending Node,
The Angle measured from reference direction, x in the reference plane to ascending node.
• Defined to be 0 ≤ ≤ 360
• Undefined for equatorial orbits (i = 0).
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The Orbital Plane Right Ascension of Ascending Node,
RAAN can be found from the line of nodes as
cos() = x · ~n
Must resolve quadrant ambiguity.
Quadrant Ambiguity: Calculators assume is in quadrant 1 or 2. Correct as
=
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Argument of Periapse, ω
• Undefined for Circular Orbits.
Definition 5.
The Argument of Periapse is the angle from line of nodes to the point of periapse.
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Argument of Periapse, ω
Can be calculated from
Must resolve quadrant ambiguity
Quadrant Ambiguity: Calculators assume ω is in quadrant 1 or 2. Correct as
ω =
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Summary: Visualization
orbitalelements.mp4
Example: Finding Orbital Elements
Problem: Suppose we observe an object in the ECI frame at position
~r =


Determine the orbital elements.
Solution: Although not necessary, as per your homework, lets first convert to canonical units (1ER = 6378.14km, 1TU = 806.3s).
~r′ = ~r


First, lets construct angular momentum, ~h, the line of nodes, ~n and the eccentricity vector, ~e.
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Example: Finding Orbital Elements Continued
We construct ~h, ~n and ~e.
~h = ~r × ~v =
~n =
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Example: Finding Orbital Elements Continued
Now we begin solving for orbital elements.
e = ~e = .8328
E = v2
r = −.088
a = − µ
2E = 5.664ER
p = h2
µ = 1.735ER
We can now calculate our three new orbital elements as indicated. Start with inclination
i = cos−1
= 87.9 deg
No quadrant ambiguity by definition. M. Peet Lecture 6: Spacecraft Dynamics 25 / 29
Example: Finding Orbital Elements Continued
Continue with RAAN, we want the angle between x and ~n.
= cos−1
= ±132.10 deg


· ~n = −.9767 < 0
Therefore, ~n is in the third quadrant, and we need to correct
= 360− 132.10 = 227.9 deg
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Example: Finding Orbital Elements Continued
Next, the argument of perigee is the angle between ~e and ~n.
ω = cos−1


ω = 53.4 deg
Example: Finding Orbital Elements Continued
Finally, we solve for true anomaly. But this is simply the angle between ~r and ~e, so we can use
f = cos−1
~r · ~v > 0
f = 92.3 deg
Summary
The Orbital Plane
• The line of nodes
• How to construct all orbital elements from ~r and ~v
• A Numerical Illustration
Spacecraft Dynamics